1. A parallel plate capacitor is filled with a linear dielectric whose dielectric constant varies linearly from the grounded bottom plate to the upper plate located at a distance d, by the relation . If the upper plate is at a potential of , obtain expressions for (a) the potential as a function of z, (b) the electric field in the dielectric and (b) the polarization charge densities on the plates.
We start with an expression for the D field in the dielectric. As the plates contain free charge densities, the D field has a strength directed from the positive plate to the negative plate. The electric field is then given by One can calculate the potential at any point by evaluating the line integral of the electric field from the bottom plate to a point at a distance z,
Substituting , we get
The expression for the electric field can be obtained by taking the negative gradient of the potential or by expressing the previously obtained expression in terms of given constants,
Thus the displacement vector is given by
Which is of course constant. The polarization vector is given by
The polarization charge densities on the plates is given by the normal component of the polarization vector. On the upper plate, z=d
On the lower plate z=0,
2. The space between the plates of a parallel plate capacitor is filled with two slabs of dielectric material. The dielectric filling the space between the grounded bottom plate at z=0 up to z=d/2 is of dielectric constant while that filling the space from z=d/2 up to the upper plate at z=d which is maintained at a constant potential is of dielectric constant . Find an expression for the potential between the plates and the polarization charge density on the two plates.
Let the charge on the plates be and the area of the plates be A. The displacement vector is D=Q/A. The electric field from z=0 to z= d/2 is and that in the region from z=d/2 to z=d is . The potential in the region 1 is
The potential at z=d/2 is . The potential in the region 2 is
The potential at z=d is then given by
where . Thus,
The charge density is given by the normal component of the polarization vector,
. If the upper plate is at a potential of 
, obtain expressions for (a) the potential as a function of z, (b) the electric field in the dielectric and (b) the polarization charge densities on the plates. 
directed from the positive plate to the negative plate. The electric field is then given by 
One can calculate the potential at any point by evaluating the line integral of the electric field from the bottom plate to a point at a distance z,
, we get
while that filling the space from z=d/2 up to the upper plate at z=d which is maintained at a constant potential 
is of dielectric constant 
. Find an expression for the potential between the plates and the polarization charge density on the two plates.
and the area of the plates be A. The displacement vector is D=Q/A. The electric field from z=0 to z= d/2 is 
and that in the region from z=d/2 to z=d is 
. The potential in the region 1 is
. The potential in the region 2 is
. Thus,
